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Today's problem comes from the 2018 Raytheon Mathcounts. An eighth grader solved this within a few seconds. It took me much longer.

There is a set of seven numbers. The average of the highest four is 10. The average of the lowest four is 5. What is the lowest possible sum of all seven numbers?

To tackle this problem, let's convert each number into a variable- a,b,c,d,e,f,g - ranked from highest to lowest.

Then the average of the first four is 10. Therefore a+b+c+d= 40.

The average of the last four is 5. Therefore d+e+f+g= 20.

If we add both numbers we get 60=a+b+c+2d+e+f+g.

Now we need to find a value of d and subtract it from 60. We are looking for the highest value for d to get the smallest sum of all numbers. What do we know about the median term? We know that when averaged with the three lowest terms, the average is five. So the smallest possible value for d is five (that is d=5, e=5, f=5,g=5). We also know that the average of the four highest terms is ten. Therefore, the largest value that d can have is 10 (that is, a=10, b=10, c=10, d=10). Thus d is greater or equal to 5 and less than or equal to 10.

Let's take d=10.

a+b+c+2d+e+f+g-d=60-10=50.

The lowest sum for the seven numbers is 50.

E-mail me if you have any questions!!

Do you have a math problem? E-mail me!!

This math problem comes from a recent Mathcounts competition.

"what is the sum of the positive integers p for which the value of 13/(p^2-3) is a positive integer?"

An eighth grade student answered this question within 30 seconds. It took me much longer.

To solve this problem, you have to realize that they are asking for cases in which p is an integer and 13/(p^2-3) is a positive integer.

Setting the denominator to zero, we find two discontinuities: 3^(1/2) and -3^(1/2). Taking the positive, let's find out if the curve 13/(p^2-3) is positive or negative as we approach 3^(1/2). BTW: 3^(1/2) is approximately 1.73.

Let's try p=1. We get 13/(p^2-3) equaling -13/3. In fact, closer we get to 1.73, the more negative this curve gets. Therefore this problem is not interested in an value of p smaller than 1.73. This problem is also not interested in any p in which 13/(p^2-3) is less than one. Solving this, we get 4.

Therefore, the values of p of interest (again, only integers) are 2, 3, and 4. Plugging in 2, we get 13, this is an integer. Plugging in 3 we get 13/6, which is not an integer. Plugging in 4, like we did above, we get 1.

To solve this problem, we then go: 4+2=6.

The answer is six. Not quite 30 seconds, but not too shabby.

Dear math students, I'm sorry I haven't posted in a long time. I just relocated to Pennsylvania, where I'll resume teaching math. If you have a home work problem or a math problem you need help solving, please e-mail them to me!!

More math problems to follow!! Stay tuned!!

Today we will expand on the same truck problem we’ve been working on the past two days.

Let’s now say that you are a manager of these truck fleets. You need to ensure that your trucks operate 198hours, but you want to use the least number of trucks possible. How big are your two fleets?

If we use the numerical methodology from two days ago, we would have to iterate this problem several times before arriving at the solution. However, if we used the graphical approach from yesterday, we can easily find the solution by looking at the excel chart that I posted.

Looking at yesterday’s chart, it’s evident that the smallest possible number of trucks is 25 trucks with 24 trucks in the first fleet and one truck in the second fleet. Let’s see if this checks out numerically using the time relationship from two days ago:
198=8x+6y
198-8(24)+6(1)
198=192+6=198
It checks out!!

In fact this solution requires four fewer trucks than our solution from two days ago (where x=12 and y=17 for a grand total of 29 trucks).

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This math problem was first brought to my attention from my dear friend Michael Allen Potter. In this post, I will solve the problem.

"A delivery firm uses one fleet of trucks on daily routes of 8 hours. A second fleet, with five more trucks than the first, is used on daily routes of 6 hours. Budget allotments allow for 198 hours of daily delivery time. How many trucks are in each fleet."

To solve this problem, we will:
1. Identify variables.
2. Identify the equations.
3. Substitute.
4. Solve.

1. Identify variables: There are two types of information in this problem: the number of trucks in either fleet and the number of total hours- which we know is 198. For this problem, let x be the number of trucks in the first fleet. Let y be the number of trucks in the second fleet.

2. Identify the right equations: We aren't given much information on how many are in the first fleet. That's okay. Let's just set up a reflexive statement:
x = x

We do have a little more information about the number of trucks in the second fleet. It's five more than what ever the first fleet is. Therefore:
y=x+5

We also need an equation for the number of hours and how this relates back to number of trucks in either fleet:
198=8x+6y

What the above is saying is that the total number of trucks in the first fleet (x) times the number hours each truck works (8hr) is the total hours the first fleet works. Similarly, the trucks in the second fleet (y) times the hours they work (6hr) is the total hours the second fleet works. The sum of these hours should equal exactly 198hrs, as described in the problem.

3. Substitute: Look carefully at the third equation above. We appear to have two variables an only one equation. That's okay!! This is a common situation. We know how x and y are related (by the second equation above). Now we need to substitute the second equation into the third:

198=8x+6(x+5)

4. Solve: Now we just plug and chug.

198=8x+6(x+5)
198=8x+6x+30
198=14x+30
168=14x
x=12

and since we know that y=x+5=12+5=17.

Therefore, your answer is 12 trucks in the first fleet and 17 trucks in the second fleet.

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